Formula for power given volumetric flow and pressure

[GUN METAL]

i need some help.........as usual !

What i am trying to do is determine what volumetric flow rate of air is required at a given pressure to calculate the power available IN THE REAL WORLD.

I HAD SOME GREAT REFERENCE ON THIS BUT I HAVE LOST THEM ALL

Obviously every engine is different with varying efficiencies etc, but for a given air flow with the right fuelling you should get a given amount of power.

In boosted engines the same volumetric flow after the blower gives more power because the air is more dense, so ither the formula needs to be based on volumetric flow at a given pressure OR mass flow rate.

i would prefer volumetric flow at a given pressure so i can relate them to some supercharger curves i have been looking at, namely the one below i am looking at installing on my 4AGZE

http://www.opcon.se/www/files/lyshol..._lys1200ax.pdf

There is some other things from the above graph i dont understand

1) presumably the flow rate on the bottom of the chart is flow rate before the blowwer ?

2) The pressure ratio is that above atmospheric so a PR of 2 is about 15 PSI ? something tells me this is NOT the resulting boost you will get as there is no allowance for different engine displacements / volumetric efficiencies etc.


3) So how do i calculate how much power this thing will make at various flows / pressures ?

this is supposedly rated at like 400 HP, and like turbos they all have a rating, just trying to work out what it is for this one


Thanks

[oldcorollas]

you can do it simply, by assuming a max of around 100Nm/L at roughly 90-100% VE, then multiplying by displacement, and by absolute pressure (ie, 2 bar boost = x3)

then you can fudgefactor it for VE

yup, x-axis should be inlet volume
that PR=2 means it is only efficient up to about 14psi boost.

[Supra967]

Don't think you can get 'real world' on the calculator. You should be able find a mass flow calculation online. From there just use

density = press[kPa]/(0.287x(temp[C]+273)

and

volume flow = mass flow/density

I use turbocalc for quick calculations:

http://www.turbofast.com.au/freesoftware.html

Using that you should be able to guess at a volume flow rate based on your engine size, expected VE and boost level

[AdrianS]

You've got too many unknowns. It' easier to pick an engine RPM for peak power/torque, and pick a blower ratio for the boost pressure you're after. Then you can read off airflow, outlet temp, etc off the graph, and calculate power from the air mass flow. I'll try to dig up the formulas, but oldcorolla's numbers look about right. Don't forget the blower drive (green lines) : 20+ kW.

[Duk]

i need some help.........as usual !

What i am trying to do is determine what volumetric flow rate of air is required at a given pressure to calculate the power available IN THE REAL WORLD.

I HAD SOME GREAT REFERENCE ON THIS BUT I HAVE LOST THEM ALL

Obviously every engine is different with varying efficiencies etc, but for a given air flow with the right fuelling you should get a given amount of power.

In boosted engines the same volumetric flow after the blower gives more power because the air is more dense, so ither the formula needs to be based on volumetric flow at a given pressure OR mass flow rate.

i would prefer volumetric flow at a given pressure so i can relate them to some supercharger curves i have been looking at, namely the one below i am looking at installing on my 4AGZE

http://www.opcon.se/www/files/lyshol..._lys1200ax.pdf

There is some other things from the above graph i dont understand

1) presumably the flow rate on the bottom of the chart is flow rate before the blowwer ?

2) The pressure ratio is that above atmospheric so a PR of 2 is about 15 PSI ? something tells me this is NOT the resulting boost you will get as there is no allowance for different engine displacements / volumetric efficiencies etc.


3) So how do i calculate how much power this thing will make at various flows / pressures ?

this is supposedly rated at like 400 HP, and like turbos they all have a rating, just trying to work out what it is for this one


Thanks

I have read the 11-12 pounds of air per minute for 100hp a few times.
That's 5-5.45kg/min. But what sort of parasitic losses (like the power to drive a supercharger) that allows for is any ones guess.

I'm guessing you're investigating using the Lysholm on a 4AGZE?

[GUN METAL]

Thanks for the replies kids !


oldcorollas, sorry but i dont understand how i use than formula to calculate power from the volumetric flow ? I have seen that calc before but i dont think it helps me.

What i need is a relationship between power and airflow, or if you take the pressure out of the equation the mass flow rate.


Basically what i am saying is if i want say 200 kw at the wheels MAX then you need a given mass of air to make that power with a given mass of fuel at the given A/F ratio.

Obviously there are losses heat / driveline etc but these are all generic factors(more or less) i just want to get into the ballpark.

An old school example of what i am getting at is with old carbies. They were /are rated at a certain CFM Cubic Feet per Minute of air, and the old school guys knew (all things being equal) you whack on a given size carby, get the AF mixtures right and you will end up with "X" HP.

So thats essentially what i am after how much air do i need to make 200 kw at the wheels (say)

ok thanks for confirming the X axis is inlet volumetric flow, ie at atmospheric


How do you figure PR of 2 is only efficient up to 14 PSI ? accoding to the curve its highest volumetric efficiy is 95 and thats right down the bottom right hand corner, but at the top right at PR of 2.2 its still 92% and its adiabatic efficieny is still high at 62ish %

Let me put this a different way, can you plot where on this graph this would work over the rev range on a 4AGZE or at least on peak power ?



Supra967 thanks for the link once i can get it running i willsee if ican make sense of it.

AdrianS, ok heres some numbers but i am not sure how it helps, feel free to plug em in and let me know.

Lets pick 8000 RMP for peak power

I was thinking 17 PSI as a maximum

not sure what you mean by blower ratio but if this thing is good for 16000 lets make the crankshaft to blower ratio 2:1 ie 8000 crank 16000 SC

I have no idea how that helps use this graph tho.

Would appreciate if you can fill in some gaps.


Duk, yep thats more along the lines i was after, and yep for the 4AGZE, hoping to replace the SC14 with something like this and get about 170 kw at the wheels at about 17 PSI. The SC12 is making 140 at the wheels on 12 PSI so 170 is hopefully conservative considdering how poor the efficiency of the SC12 is at those boost pressure / speed

Thanks

[Radar]

I have no idea if this is right but I think this is what old corolla meant?

So if your NA 4A makes 90nm/L at 100% VE
Then its 90 x 1.6 (motor size) x 2.15 (Absolute pressure: [14.7psi Atmo + 17psi Boost / 14.7 Atmo)
310 nm x 7200 peak power rpm = 2232000 / 5252 = 424hp
Then you have to take into account how much power the supercharger is taking. So assuming it takes about 20kw to drive that mean's you'll end up with 300kw?
Being supercharged for every bar of boost you increase the VE% 100% due to no restrictions in the exhaust so its easy to make loads of torque without much boost.
Turbo's suffer from a turbo being in the exhaust track so the VE% suffers to some degree.
I have read somewhere that the torque line follows the VE% line, In other words the higher the VE% the more torque you'll make...
I doubt I got it right but eh

[RA35GT]

PR = pressure ratio

PR = (Gauge (Boost) Pressure + Atmospheric pressue) / (Atmospheric pressure)

or

PR = (Absolute Pressure) / (Atmosperic Pressure)

for a PR of 2. Boost pressure is 14.7psi.

[oldcorollas]

thats the basic idea..

kW = (engine size x 100 x (1+ boost) x rpm) / 9402
kW = (1.6x90x2.156x8000)/9402
power = 264.2kW or 354hp (and 310Nm)

but then for airflow, 90%VE means 90%VE (well.. sort of.. but lets run with it)

ie, 8000rpm /2 x 1.6L = 6400L/min for NA and 100% VE
then multiply by 2.156 * 0.9 = 12421L/min
1L/min = 0.0353 CFM or 1L/min = 0.001 m^3
= 438CFM or 12.421m^3/min (for 17psi and 8000rpm)

also (for turbo) http://www.turbobygarrett.com/turbob...o_tech103.html
they say
" As a very general rule, turbocharged gasoline engines will generate 9.5-10.5 horsepower (as measured at the flywheel) for each lb/min of airflow."
and
"Air density at sea level is 0.076lb/ft3"
438CFM = 33.3lb/min, which predicts 316-350hp....

so... comparing airflow calc to the basic power calc of 354hp.. is pretty bloody close (since we are not considering losses, and turbo engine have losses as well, the calc from pure VE will be a little optimistic at higher pressures, since VE decreases as pressure goes up)

both will scale up and down for VE at the same rate.

ie assuming you have 95% VE instead of 90, then you have
13.1 m^3/min (max) and 373hp/ 279kW

so from the graph, at 13m^3/min and 2.17 pressure ratio, discharge temp will be 135C, and it will need 31-32kw to drive... (= rough net power of 246kw at flywheel)
also means a blower speed of about 12300..

call it 12500, for crank speed of 8000 means you need a pulley ratio of 1.736.

that make more sense?

[Duk]

......rough net power of 246kw at flywheel

And if you had that from a positive displacement supercharged 4AGZE, you would have 1 little engine with a bloody big pair of nuts attached to it .

[oldcorollas]

yeah, except the blower is getting out of its efficiency range , and would need very good intercooling etc

chances are it will be lower in power due to various losses... but SC 17-18psi would feel a lot stronger in midrange than turbo... and a lot more kick in part throttle

now if the blower span a little slower and was being fed some positive pressure air.......

[Duk]

Better to have a road car engine have peak PD SC efficiency lower than the peak engine speed. That way you get to use peak compressor efficiency more often than just up near the rev limiter.
Smaller PD SC geared higher makes for a broader torque curve than a bigger PD SC geared slower (to achieve the same boost pressure).


Note that I am only referring to a road car. And I am only regurgitating what I have read, but it makes excellent sense to me.

[GUN METAL]

Duk, well thats the plan !

oldcorollas, thanks for the detailed reply and yes that does make more sense, i need to sit down and go through the numbers but i suppose the result looks about right, but sadly hardly seems to be worth the effort to install one.

Maybe the numbers on paper dont stack up as well as they might in the real world. Unfortunately anyone with experience with these holds it close to their chests, i am stuffed if i can find anyone with practiacle application on these who wants to detail it.

out of interest i have just plotted the info on the next size up and there is barely any improvemnt in efficieny or lower outlet temps and the power to drive it is much the same at 30 odd kw.

http://www.opcon.se/www/files/lyshol..._lys1600ax.pdf

Maybe its a simple fact that these are to be used in engines with larger capacities and less boost ie about 10 psi or thereabouts ? sems the larger you go it simply enables more flow not pressure without it going off the scale ?

The thing is tho i was of the understanding these are so much more efficient than roots type blowers but maybe they are not ? probly just the usual sales blurb. even spinning the SC12 to around 12000 RPM i still couldnt get the discharge temps over 120 degrees and that was on a 40 degree day, it just doesnt stack up, if they are so efficient why do they get so hot ?

What sort of blower were they using on the TURD Aurion if i beleive what they tell me they didnt even have an intercooler on it, actually neither did the TRD Hilux or am i dreaming ?

The weird thing is that the engine that i know they used these on from favtory the Eunos 800 they supposeldy ran them at like 14 psi from the factory and still only made 160 kw at the flywheel ??????

http://en.wikipedia.org/wiki/Mazda_K_engine

http://autospeed.com/cms/title_Impor...4/article.html



The thing is that not relfected in these graphs is how if you whack a big charger on a small engine you would think you would get more boost, ie its trying to pump in more air but cant so it increases the pressure. Like when you swap an SC12 for a 14, you run it at the same speed and all of a sudden you get a stack more boost.

The other thing i have noticed after looking at a lot of kW ratings of jap engines copared to actual chassis dynos is the ratio of the losses.

a 200kW engine dynod at 130 kW

a 150 kW engine dynod at 111 kW

a 88 kw engine dynod at 60 kW

These are only a few and all on different dynos but at least its an indicator.

[Duk]

Your outright speed will come from having a higher average torque curve rather than just massive outright numbers.
If OC's calculated (post SC drive loss) power figures are reasonably accurate, that kind of power in an AW11 would be an absolute hoot, because it (should) have some damn good average torque.
With an excellent intercooler and maybe some post SC water injection, you should be able to squeeze some decent high load ignition advance angles into the thing and chase the top HP figures.

Remember that the Miller cycle engine uses the SC to help push exhaust gasses out of the combustion chamber and the concept is more about fuel economy and emissions than outright power. The dyno curve would make interesting viewing.

[GUN METAL]

Duk, check the Autospeed link the curve is in there


hey old maybe this answers my Q check out post 6

http://www.toymods.org.au/forums/showthread.php?t=1738